列表索引超出范围：两个列表的迭代

I'm having difficulty with printing text from one list, and playing the soundbyte for each list after it prints. However if there is a second sentence, but no second soundbyte, I get list index out of range.

我很难从一个列表中打印文本，并在打印后为每个列表播放声音字节。但是，如果有第二个句子，但没有第二个声音，我得到列表索引超出范围。

What do I do if the second list has no second item? etc...

如果第二个列表没有第二个项目，我该怎么办？等等...

import pygame
from pygame.locals import *
pygame.mixer.init()
a="first sentence..."
b="second sentence..."
sentList=[]
sentList.append(a)
sentList.append(b)
sound1= pygame.mixer.Sound('lose1.ogg')
soundList=[]
soundList.append(sound1)
chan1=""
for i in range (len(sentList)):
    print sentList[i]
    chan1= soundList[i].play()
    while chan1.get_busy():
        z=0

3 个解决方案

#1

you can use zip, a simple example:

你可以使用zip，一个简单的例子：

>>> a = [1,2,3]
>>> b = [4,5]
>>> for i,j in zip(a, b):
        print i, j
1 4
2 5

so your code can be like this:

所以你的代码可以是这样的：

for i,j in zip(sentList, soundList):
    print i
    chan1 = j.play()
    while chan1.get_busy():
        z = 0

#2

Ok -

好 -

sentList has a length of 2 (you append it twice). soundList has a length 1. In your for loop, you are trying to access the second element of soundList, which does't exist. That is why an error is thrown. To solve this, you can enclose chan1= soundList[i].play() in a try catch to look like this.

sentList的长度为2（您将其追加两次）。 soundList的长度为1.在for循环中，您尝试访问soundList的第二个元素，该元素不存在。这就是抛出错误的原因。要解决这个问题，你可以在try catch中包含chan1 = soundList [i] .play（），看起来像这样。

try:
    chan1= soundList[i].play()
except:
    print("error")

#3

you can put a condition in to see if the item exists in the list:

你可以放入一个条件来查看列表中是否存在该项：

if len(soundList) > i:
   chan1= soundList[i].play()
   while chan1.get_busy():
      z=0

or you use a try block:

或者您使用try块：

try:
   chan1= soundList[i].play()
   while chan1.get_busy():
      z=0
except IndexError:
    print 'ListIndex does not exist'

#1